## Understanding a household of curves and their trajectories

Given a household of curves, prefer ???y=kx???, us can pick different values for ???k??? to create the equations of some of the curve in the family.

You are watching: Give the family of orthogonal trajectories of

Each the the values of ???k??? above give a various curve the is part of the household of curves provided by ???y=kx???. If we graph every of the curves over together on the very same graph, us get

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Sometimes you’ll be asked to find the orthogonal trajectories come a family members of curves. The orthogonal trajectories room the curves that room perpendicular to the family members everywhere. In various other words, the orthogonal trajectories are one more family of curves in which every curve is perpendicular come the curves in original family. In the example below we’ll display how to use calculus to find the orthogonal trajectories, however for now we’ll give away the price so the we deserve to sketch the family of orthogonal trajectories and also see the they are perpendicular to the original family.

The curve in blue are from the original family ???y=kx???. The curve in environment-friendly are

???x^2+y^2=1???

???x^2+y^2=2???

???x^2+y^2=3???

???x^2+y^2=4???

which are four of their orthogonal trajectories, part of their totality family the orthogonal trajectories given by ???x^2+y^2=C???. An alert how each green circle is perpendicular come every blue line, where a green and blue curve intersect one another.

Now that we have actually an idea the what we’re trying to find, let’s shot an example where we use calculus to show that the orthogonal trajectories are provided by ???x^2+y^2=C???.

The orthogonal trajectories are the collection of curves that are always perpendicular come a particular family the curves

## Step-by-step example of just how to find the orthogonal trajectories

Example

Find the orthogonal trajectories to the family of curves.

???y=kx???

We constantly start by using implicit differentiation to take it the derivative the both sides, and also then we’ll deal with for ???dy/dx???.

???(1)fracdydx=k(1)???

???fracdydx=k???

Once we’ve got an equation for ???dy/dx???, fine go ago to the initial equation and solve it because that ???k???, so the we deserve to plug a value in because that ???k??? it is in terms of ???x??? and also ???y???.

???y=kx???

???k=fracyx???

Plug the value for ???k??? into the equation because that ???dy/dx???.

???fracdydx=k???

???fracdydx=fracyx???

Remember that ???dy/dx??? is the slope of the family of curve ???y=kx???, so the equation we just discovered represents the slope of the household everywhere.

If we desire to find the orthogonal trajectories, and also we recognize that castle perpendicular come our family everywhere, climate we desire a slope for the orthogonal trajectories that is perpendicular come the steep of the original family. To find a perpendicular slope, we take the negative reciprocal (flip that upside down and add a negative sign).

Slope that the original family: ???fracdydx=fracyx???

Slope that the orthogonal trajectories: ???fracdydx=-fracxy???

If we treat the slope of the orthogonal trajectories together a separable differential equation, we have the right to separate variables and integrate both political parties in bespeak to uncover the equation that the household of orthogonal trajectories. We’ll begin by separating variables.

???fracdydx=-fracxy???

???dy=-fracxy dx???

???y dy=-x dx???

Then we’ll combine both sides.

???int y dy=int -x dx???

???frac12 y^2=-frac12 x^2+C???

???frac12 x^2+frac12 y^2=C???

???x^2+y^2=2C???

The ???2??? can be absorbed into the constant, therefore the equation becomes

???x^2+y^2=C???

This is the equation of the household of orthogonal trajectories, which is the family of curve that room perpendicular come the initial family.